Solve for $x$ : $5x^2 - 20x + 20 = 0$
Dividing both sides by $5$ gives: $ x^2 {-4}x + {4} = 0 $ The coefficient on the $x$ term is $-4$ and the constant term is $4$ , so we need to find two numbers that add up to $-4$ and multiply to $4$ The number $-2$ used twice satisfies both conditions: $ {-2} + {-2} = {-4} $ $ {-2} \times {-2} = {4} $ So $(x - {2})^2 = 0$ $x - 2 = 0$ Thus, $x = 2$ is the solution.